Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $t = \dfrac{3z - 27}{3z^2 + 3z - 168} \times \dfrac{z^2 + 5z - 24}{z - 3} $
Explanation: First factor out any common factors. $t = \dfrac{3(z - 9)}{3(z^2 + z - 56)} \times \dfrac{z^2 + 5z - 24}{z - 3} $ Then factor the quadratic expressions. $t = \dfrac {3(z - 9)} {3(z + 8)(z - 7)} \times \dfrac {(z + 8)(z - 3)} {z - 3} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac {3(z - 9) \times (z + 8)(z - 3) } { 3(z + 8)(z - 7) \times (z - 3)} $ $t = \dfrac {3(z + 8)(z - 3)(z - 9)} {3(z + 8)(z - 7)(z - 3)} $ Notice that $(z + 8)$ and $(z - 3)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {3\cancel{(z + 8)}(z - 3)(z - 9)} {3\cancel{(z + 8)}(z - 7)(z - 3)} $ We are dividing by $z + 8$ , so $z + 8 \neq 0$ Therefore, $z \neq -8$ $t = \dfrac {3\cancel{(z + 8)}\cancel{(z - 3)}(z - 9)} {3\cancel{(z + 8)}(z - 7)\cancel{(z - 3)}} $ We are dividing by $z - 3$ , so $z - 3 \neq 0$ Therefore, $z \neq 3$ $t = \dfrac {3(z - 9)} {3(z - 7)} $ $ t = \dfrac{z - 9}{z - 7}; z \neq -8; z \neq 3 $